Which of the following describes the sampling distribution of the sample mean, given that the daily revenue X at a university snack bar has a mean of 4400 and a variance of 400, an... Which of the following describes the sampling distribution of the sample mean, given that the daily revenue X at a university snack bar has a mean of 4400 and a variance of 400, and 150 days are randomly selected?
Understand the Problem
The question is asking us to determine the sampling distribution of the sample mean given the population mean, variance, and sample size. We need to use the Central Limit Theorem to determine the distribution, mean, and variance of the sample mean.
Answer
$\bar{X} \sim N(100, 0.25)$
Answer for screen readers
The sampling distribution of the sample mean is approximately normal with a mean of 100 and a variance of 0.25. This can be written as $\bar{X} \sim N(100, 0.25)$.
Steps to Solve
- Identify the population parameters
The problem states that the population mean is $\mu = 100$ and the population variance is $\sigma^2 = 25$. Therefore, the population standard deviation is $\sigma = \sqrt{25} = 5$.
- Identify the sample size
We are given that the sample size is $n = 100$.
- Apply the Central Limit Theorem (CLT)
The CLT states that the sampling distribution of the sample mean, $\bar{X}$, approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution. The mean of the sampling distribution is equal to the population mean, and the variance of the sampling distribution (also called the standard error) is equal to the population variance divided by the sample size.
- Determine the mean of the sampling distribution
The mean of the sampling distribution of the sample mean is: $$ \mu_{\bar{X}} = \mu = 100 $$
- Determine the variance of the sampling distribution
The variance of the sampling distribution of the sample mean is: $$ \sigma_{\bar{X}}^2 = \frac{\sigma^2}{n} = \frac{25}{100} = 0.25 $$
- Determine the standard deviation of the sampling distribution
The standard deviation of the sampling distribution of the sample mean (also known as the standard error) is: $$ \sigma_{\bar{X}} = \sqrt{\sigma_{\bar{X}}^2} = \sqrt{0.25} = 0.5 $$
- State the sampling distribution
The sampling distribution of the sample mean is approximately normal with a mean of 100 and a variance of 0.25 (or standard deviation of 0.5). We write this as: $\bar{X} \sim N(100, 0.25)$.
The sampling distribution of the sample mean is approximately normal with a mean of 100 and a variance of 0.25. This can be written as $\bar{X} \sim N(100, 0.25)$.
More Information
The Central Limit Theorem is a fundamental concept in statistics because it allows us to make inferences about a population based on a sample, even if we don't know the exact distribution of the population. As the sample size increases, the sampling distribution of the sample mean becomes more and more normal, regardless of the population's distribution.
Tips
A common mistake is to forget to divide the population variance by the sample size when calculating the variance of the sampling distribution. Another mistake is to confuse the standard deviation of the population with the standard deviation of the sampling distribution (standard error). Remember to take the square root of the variance to get the standard deviation.
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